Divisibility by 11: Proving the Alternating Sum Method

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SUMMARY

The divisibility rule for 11 states that a number is divisible by 11 if the alternating sum of its digits is divisible by 11. This can be proven using a two-digit number represented as n = a*10 + b, where a is the first digit and b is the second digit. By rewriting n as n = a*10 + (b - a) + a, it is established that a*10 + a is divisible by 11. Consequently, if (b - a) is also divisible by 11, then n is divisible by 11.

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Homework Statement



what is the test to to see if a number is divisible by 11 and prove it.


The Attempt at a Solution



If the alternating sum of a numbers digits is divisble by 11 then so is the number.


I don't know how to prove it tho.
 
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Do a simple case. Take a two digit number n. First digit a, second digit b. So n=a*10+b. Rewrite this as n=a*10+((b-a)+a)=a*10+a+(b-a). Now a*10+a is always divisible by 11 (why?). So if (b-a) is divisible by 11, then n is.
 
because it equals 11a.

thanks!
 

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